Theorem: Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then it has a limit point.

So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=1$ we constructed a sequence of intervals, such that each iteration has infinite number of points of the set, but what is the anlogy in $\mathbb{R}^l$?, Can someone help me to prove this please?

Thanks a lot in advance :)


You can construct a sequence of hypercubes instead of intervals.

For example, taking $l=2$, lets assume (without loss of generality) that $A\subseteq [0,1]^2$.

Then, split $[0,1]^2$ into four subsquares (a square is a $2D$ cube):

$$A_{11} = [0,1/2]\times [0,1/2]\\ A_{12} = [1/2, 1]\times [0,1/2]\\ A_{21} = [0,1/2]\times [1/2, 1]\\ A_{22} = [1/2, 1]\times [1/2,1]$$

You know that one of the squares has infinitely many points. Pick that square and mark it $A_1$ and split it again.

In this way, you can construct a series of squares $$A_1\supseteq A_2\supseteq A_3\supseteq\cdots$$

And you can show that the intersection of these squares contains one point $x$. And that point must be a limit point of $A$, because it lies in $A_k$ and infinitely many other points also lie in $A_k$, so infinitely many points are at most $\frac{1}{2^k}$ away from $x$.

  • $\begingroup$ Ok let me check, If I get stuck, Can I let you know?, just give a while to think in your idea :) $\endgroup$ – user162343 Sep 3 '15 at 15:47
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    $\begingroup$ @user162343 Of course. Anytime you have a question, drop a comment under this answer so I get a notification. $\endgroup$ – 5xum Sep 3 '15 at 15:49

The closure of $A$ is compact. In compact metric spaces each sequence has a convergent subsequence. If you may use that result you are done.

Elaboration: Since $A$ is infinite there is a injective sequence $(a_n)_{n\in\mathbb N}\subseteq A$. Since $\overline{A}$ is compact $(a_n)_{n\in\mathbb N}$ has a convergent subsequence with limit in $\overline{A}$. Since the limit is the limit of an injective sequence of elements of $A$ it is a limit point of $A$.

  • $\begingroup$ Can you elaborate more in this answer please? $\endgroup$ – user162343 Sep 3 '15 at 15:48
  • $\begingroup$ @user162343 Do you mean how you get the limit point? $\endgroup$ – principal-ideal-domain Sep 3 '15 at 15:48
  • $\begingroup$ Well we need a candidate right?, and yes I want to show at least a limit point :) May be by construction or other means :) $\endgroup$ – user162343 Sep 3 '15 at 15:50
  • $\begingroup$ @user162343 Better now? $\endgroup$ – principal-ideal-domain Sep 3 '15 at 15:53
  • $\begingroup$ Right :), the only thing I dont understand is the injective part :), how do you know it is injective? $\endgroup$ – user162343 Sep 3 '15 at 15:54

Proving it for $l=1$ is enough to get the rest.

Let $X_n = (x_{1,n},\dots,x_{l,n})$ be an injective infinite sequence of points in the set $A$.

$x_{1,n}$ is bounded

So there is a subsequence s.t $x_{1,\phi_1(n)}$ converges

Now consider $X_{\phi_1(n)} = (x_{1,\phi_1(n)},\dots,x_{l,\phi_1(n)})$

$x_{2,\phi_1(n)}$ is bounded

So there is a subsequence s.t $x_{2,\phi_1\circ\phi_2(n)}$ converges

(Note that $x_{1,\phi_1\circ\phi_2(n)}$ also converges)

We repeat that until we find a subsequence that makes all the coordinates converge.

$(x_{1,\phi_1\circ\phi_2\dots\circ\phi_l(n)},\dots,x_{l,\phi_1\circ\phi_2\dots\circ\phi_l(n)})$ all converge, so $X_{\phi_1\circ\phi_2\dots\circ\phi_l(n)}$ also converges and its limit is a limit point of $A$

  • $\begingroup$ Thanks a lot for clarifying but who is the sequence $x_{2}$? $\endgroup$ – user162343 Sep 3 '15 at 23:12
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    $\begingroup$ @user162343 $x_{1,n},\dots,x_{l,n}$ are the coordinates of $X_n$ $\endgroup$ – Kitegi Sep 4 '15 at 17:01

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